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Are the real numbers the unique Dedekind-complete ordered set?
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A totally ordered set is Dedekind-complete if any subset which has an upper bound also has a least upper bound. Now any two ordered fields which are Dedekind-complete are order-isomorphic as well as field-isomorphic. Another way of phrasing this is that the second-order theory of real numbers is categorical, i.e. it has a unique model upto isomorphism.
But I’m interested in what happens if you drop the field structure. My question is, are any two dense, totally ordered, Dedekind-complete sets order-isomorphic? In other words, is the real number system the unique dense totally ordered Dedekind complete set?
abstract-algebra logic field-theory order-theory real-numbers
asked 5 hours ago
Keshav Srinivasan
1,81111339
add a comment |Â
up vote
2
down vote
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A totally ordered set is Dedekind-complete if any subset which has an upper bound also has a least upper bound. Now any two ordered fields which are Dedekind-complete are order-isomorphic as well as field-isomorphic. Another way of phrasing this is that the second-order theory of real numbers is categorical, i.e. it has a unique model upto isomorphism.
But I’m interested in what happens if you drop the field structure. My question is, are any two dense, totally ordered, Dedekind-complete sets order-isomorphic? In other words, is the real number system the unique dense totally ordered Dedekind complete set?
abstract-algebra logic field-theory order-theory real-numbers
asked 5 hours ago
Keshav Srinivasan
1,81111339
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A totally ordered set is Dedekind-complete if any subset which has an upper bound also has a least upper bound. Now any two ordered fields which are Dedekind-complete are order-isomorphic as well as field-isomorphic. Another way of phrasing this is that the second-order theory of real numbers is categorical, i.e. it has a unique model upto isomorphism.
But I’m interested in what happens if you drop the field structure. My question is, are any two dense, totally ordered, Dedekind-complete sets order-isomorphic? In other words, is the real number system the unique dense totally ordered Dedekind complete set?
abstract-algebra logic field-theory order-theory real-numbers
asked 5 hours ago
Keshav Srinivasan
1,81111339
A totally ordered set is Dedekind-complete if any subset which has an upper bound also has a least upper bound. Now any two ordered fields which are Dedekind-complete are order-isomorphic as well as field-isomorphic. Another way of phrasing this is that the second-order theory of real numbers is categorical, i.e. it has a unique model upto isomorphism.
But I’m interested in what happens if you drop the field structure. My question is, are any two dense, totally ordered, Dedekind-complete sets order-isomorphic? In other words, is the real number system the unique dense totally ordered Dedekind complete set?
abstract-algebra logic field-theory order-theory real-numbers
abstract-algebra logic field-theory order-theory real-numbers
asked 5 hours ago
Keshav Srinivasan
1,81111339
asked 5 hours ago
Keshav Srinivasan
1,81111339
asked 5 hours ago
Keshav Srinivasan
1,81111339
asked 5 hours ago
Keshav Srinivasan
1,81111339
asked 5 hours ago
Keshav Srinivasan
1,81111339
1,81111339
add a comment |Â
add a comment |Â
1 Answer
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oldest
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up vote
4
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No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $mathbbR$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.
However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $mathbbR$ as the completion of $mathbbQ$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^aleph_0$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $mathbbR$ since it does not even have the same cardinality.
(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $YtimesmathbbQ$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)
Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $mathbbR$ from that of $mathbbQ$ to get a field structure on the completion of $X$.
However, it turns out that the verification of some of the field axioms on $mathbbR$ uses more than just that it is the completion of an ordered field $mathbbQ$--it also uses the fact that $mathbbQ$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $mathbbR$.
answered 5 hours ago
Eric Wofsey
168k12196313
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
2
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
1
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $mathbbR$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.
However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $mathbbR$ as the completion of $mathbbQ$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^aleph_0$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $mathbbR$ since it does not even have the same cardinality.
(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $YtimesmathbbQ$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)
Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $mathbbR$ from that of $mathbbQ$ to get a field structure on the completion of $X$.
However, it turns out that the verification of some of the field axioms on $mathbbR$ uses more than just that it is the completion of an ordered field $mathbbQ$--it also uses the fact that $mathbbQ$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $mathbbR$.
answered 5 hours ago
Eric Wofsey
168k12196313
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
2
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
1
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
add a comment |Â
up vote
4
down vote
accepted
No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $mathbbR$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.
However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $mathbbR$ as the completion of $mathbbQ$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^aleph_0$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $mathbbR$ since it does not even have the same cardinality.
(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $YtimesmathbbQ$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)
Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $mathbbR$ from that of $mathbbQ$ to get a field structure on the completion of $X$.
However, it turns out that the verification of some of the field axioms on $mathbbR$ uses more than just that it is the completion of an ordered field $mathbbQ$--it also uses the fact that $mathbbQ$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $mathbbR$.
answered 5 hours ago
Eric Wofsey
168k12196313
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
2
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
1
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $mathbbR$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.
However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $mathbbR$ as the completion of $mathbbQ$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^aleph_0$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $mathbbR$ since it does not even have the same cardinality.
(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $YtimesmathbbQ$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)
Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $mathbbR$ from that of $mathbbQ$ to get a field structure on the completion of $X$.
However, it turns out that the verification of some of the field axioms on $mathbbR$ uses more than just that it is the completion of an ordered field $mathbbQ$--it also uses the fact that $mathbbQ$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $mathbbR$.
answered 5 hours ago
Eric Wofsey
168k12196313
No, not at all. First of all, there are some trivial examples: you could take the empty set, or a singleton set, or a closed (or half-open) interval in $mathbbR$. To eliminate such trivialities, you should additional ask for your ordered set to be nonempty and have no greatest or least element.
However, there are also nontrivial examples that cannot be ruled out so easily. Indeed, every totally ordered set has a Dedekind-completion which can be constructed just like $mathbbR$ as the completion of $mathbbQ$. So for instance, if you start with any dense ordered set $X$ of cardinality greater than $2^aleph_0$ with no greatest or least element, its Dedekind-completion is a Dedekind-complete dense ordered set with no greatest or least element which cannot be isomorphic to $mathbbR$ since it does not even have the same cardinality.
(For an explicit example of such an $X$, note that if $Y$ is any totally ordered set (e.g., an ordinal), then $YtimesmathbbQ$ with the lexicographic order is a dense totally ordered set with no greatest or least element.)
Incidentally, you might wonder why you can't construct counterexamples for fields in the same way. After all, if you have any ordered field $X$, you can take its Dedekind-completion as an ordered set. You could then try to imitate the construction of the field structure on $mathbbR$ from that of $mathbbQ$ to get a field structure on the completion of $X$.
However, it turns out that the verification of some of the field axioms on $mathbbR$ uses more than just that it is the completion of an ordered field $mathbbQ$--it also uses the fact that $mathbbQ$ is Archimedean. So, you can similarly construct a completion of any Archimedean ordered field, but that will always just give you $mathbbR$.
answered 5 hours ago
Eric Wofsey
168k12196313
edited 4 hours ago
answered 5 hours ago
Eric Wofsey
168k12196313
answered 5 hours ago
Eric Wofsey
168k12196313
answered 5 hours ago
Eric Wofsey
168k12196313
168k12196313
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
2
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
1
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
add a comment |Â
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
2
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
1
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
Thanks for your answer. So does that mean that it’s not possible to put a field structure on, say, the Long Line?
– Keshav Srinivasan
4 hours ago
2
2
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
That's correct (assuming you want the field structure to be compatible with the order), because the long line is Dedekind complete but not order-isomorphic to $mathbbR$.
– Eric Wofsey
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
Your edit is interesting. Do you know what part of the proof uses the Archimedean property?
– Keshav Srinivasan
4 hours ago
1
1
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
Well, it depends exactly how you formulate the definition of the field operations. But for at least one formulation, the existence of additive inverses uses the Archimedean property. (You can get around this by defining addition using separate cases for positive and negative numbers, but then associativity of addition will require the Archimedean property.)
– Eric Wofsey
4 hours ago
add a comment |Â
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